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[tex]\large\underline{\sf{Solution-}}[/tex]

Given that,

[tex]\sf \: tan\phi + sin\phi = m \\ [/tex]

and

[tex]\sf \: tan\phi - sin\phi = n \\ [/tex]

Now, Consider

[tex]\sf \: {m}^{2} - {n}^{2} \\ [/tex]

[tex]\sf \: = \: {(tan\phi + sin\phi )}^{2} - {(tan\phi - sin\phi )}^{2} \\ [/tex]

We know,

[tex]\boxed{\sf \: {(x + y)}^{2} - {(x - y)}^{2} = 4xy \: } \\ [/tex]

So, using this identity, we get

[tex]\sf \: = \: 4tan\phi sin\phi \\ [/tex]

Thus,

[tex]\implies\sf \: {m}^{2} - {n}^{2} \: = \: 4tan\phi sin\phi \\ [/tex]

So, on squaring both sides, we get

[tex]\implies\bf \: ( {m}^{2} - {n}^{2})^{2} \: = \: 16tan^{2} \phi sin^{2} \phi - - - (1)\\ [/tex]

Now, Consider

[tex]\sf \: 16mn \\ [/tex]

[tex]\sf \: = \: 16(tan\phi + sin\phi )(tan\phi - sin\phi ) \\ [/tex]

[tex]\sf \: = \: 16(tan^{2} \phi - sin^{2} \phi ) \\ [/tex]

[tex]\left[ \because \: \sf \: (x + y)(x - y) = {x}^{2} - {y}^{2} \right] \\ [/tex]

[tex]\sf \: = \: 16\bigg( \dfrac{ {sin}^{2}\phi }{ {cos}^{2} \phi } - {sin}^{2}\phi \bigg) \\ [/tex]

[tex]\sf \: = \: 16 {sin}^{2}\phi \bigg( \dfrac{1}{ {cos}^{2} \phi } - 1\bigg) \\ [/tex]

[tex]\sf \: = \: 16 {sin}^{2}\phi \bigg( {sec}^{2} \phi - 1\bigg) \\ [/tex]

[tex]\sf \: = \: 16 {sin}^{2}\phi \bigg( {tan}^{2} \phi \bigg) \\ [/tex]

[tex]\sf \: = \: 16 {sin}^{2}\phi {tan}^{2} \phi \\ [/tex]

So,

[tex]\implies\bf \: 16mn = \: 16 {sin}^{2}\phi {tan}^{2} \phi - - - (2)\\ [/tex]

From equation (1) and (2), we concluded that

[tex]\implies\sf \: \boxed{\bf \: {( {m}^{2} - {n}^{2})}^{2} = 16 mn \: } \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional information

[tex]\begin{gathered}\: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sinx = \dfrac{1}{cosecx} }\\ \\ \bigstar \: \bf{cosx = \dfrac{1}{secx} }\\ \\ \bigstar \: \bf{tanx = \dfrac{sinx}{cosx} = \dfrac{1}{cotx} }\\ \\ \bigstar \: \bf{cot x= \dfrac{cosx}{sinx} = \dfrac{1}{tanx} }\\ \\ \bigstar \: \bf{cosec x = \dfrac{1}{sinx} }\\ \\ \bigstar \: \bf{secx = \dfrac{1}{cosx} }\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 } \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\\: \end{array} }}\end{gathered}\end{gathered}\end{gathered} \\ \\[/tex]

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